Problem: Consider the polar curve $r=3\sin(\theta)$. What is the slope of the tangent line to the curve $r$ when $\theta = \dfrac{\pi}{3}$ ? Give an exact expression. $\text{slope }=$
Explanation: The slope of the tangent line at a point is equal to $\dfrac{dy}{dx}$ at that point. In the case of polar curves, we can use the relationship: $\dfrac{dy}{dx}=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)}$ For a polar curve, $x={r}\cos(\theta)$ and $y={r}\sin(\theta)$. Therefore, in our problem we have: $\begin{aligned} x&={3\sin(\theta)}\cos(\theta) \\\\ y&={3\sin(\theta)} \sin(\theta) \\\\ &=3\sin^2(\theta) \end{aligned}$ Let's find $\dfrac{dy}{dx}$. $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)} \\\\ &=\dfrac{3\sin\left(2\theta\right)}{3\cos\left(2\theta\right)} \\\\ &=\tan\left(2\theta\right) \end{aligned}$ Finally, we evaluate $\dfrac{dy}{dx}$ at ${\theta = \dfrac{\pi}{3}}$. $\begin{aligned} \left. \dfrac{dy}{dx} \right| _{\theta =\tfrac{\pi }{3}}&=\tan\left(2\left({\dfrac{\pi}{3}}\right)\right) \\\\&=\tan\left(\dfrac{2\pi}{3}\right) \\\\ &=-\sqrt{3} \end{aligned}$ The slope of the tangent line to the curve $r$ when $\theta=\dfrac{\pi}{3}$ equals $-\sqrt{3}$. The graph of the tangent is shown. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${0}$ ${\frac{1}{6}\pi}$ ${\frac{1}{3}\pi}$ ${\frac{1}{2}\pi}$ ${\frac{2}{3}\pi}$ ${\frac{5}{6}\pi}$ ${\pi}$ ${\frac{7}{6}\pi}$ ${\frac{4}{3}\pi}$ ${\frac{3}{2}\pi}$ ${\frac{5}{3}\pi}$ ${\frac{11}{6}\pi}$